Projectile Motion: Finding Rifle Angles for a 91.4m Target

[mslena79]

Homework Statement

A rifle has been sighted in for a 91.4m target. If the muzzle speed of the bullet is v0=427 m/s, what are the two possible angles theta1 and theta2 betweent the rifle barrel and the horizontal such that the bullet will hit the target? One of the angles is so large that it is never used in target shooting.

Homework Equations

I know you have to use trig identities, but I'm not sure how to get to that point.
Maybe get the time it takes in one dimension to go half way?

The Attempt at a Solution

 

[learningphysics]

Write out the equations for vertical displacement and horizontal displacement.

 

[mslena79]

x-x0=delta x, y-y0=delta y, delta x=91.4m, delta y=0

 

[learningphysics]

x-x0=delta x, y-y0=delta y, delta x=91.4m, delta y=0

yes, write these using velocity, time, theta...

 

[mslena79]

is the v0 the same throughout the problem, or is it different for v0x, and voy?

 

[mslena79]

91.4m=vx*cos(theta)t, 0m=vy*sin(theta)t + 1/2(-9.80 m/s^2)t^2

 

[learningphysics]

91.4m=vx*cos(theta)t, 0m=vy*sin(theta)t + 1/2(-9.80 m/s^2)t^2

vy = v0sin(theta), and vx = v0cos(theta)

The equations should be:

91.4m=v0*cos(theta)t,
0m=v0*sin(theta)t + 1/2(-9.80 m/s^2)t^2

using these two equations try to solve for theta

 

[mslena79]

what do I do with the t variable? Is it the time is takes to go 91.4m at 427m/s? .214s.

 

[learningphysics]

what do I do with the t variable? Is it the time is takes to go 91.4m at 427m/s? .214s.

Solve for t in one equation... substitute into the other equation... then solve that equation for theta.

Yes, it is the time it takes to reach 91.4m... but how did you get 0.214s ?

 

[mslena79]

from v0 in a one dimensional equation.

 

[learningphysics]

from v0 in a one dimensional equation.

the time here will be different because of the angle...

use t = \frac{91.4}{v0*cos(theta)}

and substitute into the other equation...

 

[mslena79]

0m=(427m/s)*sin(theta)*((91.4m)/((427m/s)*cos(theta))+(1/2)*(-9.80m/s^2)*((91.4m)/((427m/s)*cos(theta))^2
(0.225/cos^2(theta))=(91.4*sin(theta))/cos(theta)

 

[learningphysics]

0m=(427m/s)*sin(theta)*((91.4m)/((427m/s)*cos(theta))+(1/2)*(-9.80m/s^2)*((91.4m)/((427m/s)*cos(theta))^2
(0.225/cos^2(theta))=(91.4*sin(theta))/cos(theta)

You can cancel cos(theta) from both sides... then use a trig. identity...

but one thing I should have mentioned:

0m=v0*sin(theta)t + 1/2(-9.80 m/s^2)t^2

can be simplified to

0m=v0*sin(theta) + 1/2(-9.80 m/s^2)t

by dividing both sides by t... since we aren't dealing with the t = 0 case.

So solving:

91.4m=v0*cos(theta)t
0m=v0*sin(theta) + 1/2(-9.80 m/s^2)t

will also work...

 

[mslena79]

sin(2theta)=.00492, can't remember how to get theta from here.

 

[learningphysics]

sin(2theta)=.00492, can't remember how to get theta from here.

take the inverse sin of both sides... that will give 2theta = ... use your calculator to get the inverse sin of 0.00492. the calculator will only give one value... but there are two angles that give the same sin.

Using 2theta = x (where x comes from the inverse sin), you solve and get theta = x/2


You did everything correctly. But I just wanted to also show the formula when we don't plug in the numbers right away...

R = v0*cos(theta)t (where R is the range)
0 = v0sin(theta) - (1/2)gt

0 = v0sin(thet) - (1/2)g[R/(v0cos(theta))]

0 = 2v0^2sin(theta)cos(theta) - gR

gR = v0^2 sin(2theta)

sin(2\theta) = \frac{gR}{{v_0}^2}

plugging in R and v0 here I get 0.004913